Ascending or descending property of a sequence according to the value of its first element Suppose that $(a_n)_{n≥1}$ is a sequence of real numbers satisfying $a_{n+1}=\frac{3a_n}{2+a_n}$.
Suppose $0<a_1<1$, then prove that the sequence $a_n$ is increasing and hence show that $lim_{n \rightarrow \infty}a_n=1$.
Suppose $a_1>1$, then prove that the sequence $a_n$ is decreasing and hence show that $lim_{n \rightarrow \infty}a_n=1$.
I attempted to write $a_{n+1}-a_n=\frac{a_n(1-a_n)}{2+a_n}$ but I am kind of stuck here.
 A: Consider first term = $\alpha\\$
Then second term = $\dfrac{3\alpha}{2+\alpha}\\$
Third term = $\dfrac{9\alpha}{4+5\alpha}\\$
Forth term = $\dfrac{27\alpha}{8+19\alpha}\\$
Fifth term = $\dfrac{81\alpha}{16+65\alpha}\\$
General term = $\dfrac{3^n\alpha}{2^n+(3^n-2^n)\alpha} , n =0,1,2...\\$
Now put limit n tending to infinity in the general term... you'll get 1 regardless of $\alpha$
Simplify to see that:
$f(n) = \dfrac{3^n\alpha}{3^n\alpha+2^n(1-\alpha)}\\$ If $\alpha$ is between 0 and 1 denominator is greater than numerator and if $\alpha > 1$ denominator is smaller than numerator.
Now to prove  that the general term is increasing or decreasing consider $f(n) = \dfrac{3^n\alpha}{2^n+(3^n-2^n)\alpha} \\$
Differentiate to find that:
$f'(x) = -\dfrac{(ln(3)-ln(2))\alpha(\alpha-1)2^x3^x}{(3^x\alpha+2^x(1-\alpha))^2} \\$
Note the sign $f'(x)$ is solely determined by the value of $\alpha$, because all the other terms are strictly positive. See below:
$f'(x) = -\alpha(\alpha-1) \left(\dfrac{(ln(3)-ln(2))2^x3^x}{(3^x\alpha+2^x(1-\alpha))^2} \right)\\$
We see when $\alpha \in (0,1) f'(x)>0$ and when $\alpha>1 f'(x)<0$... implies $f(x)$ is increasing when $0<\alpha<1$ and decreasing when $\alpha>1. \\$
Comment down below if any doubts.
A: Note that$$(\forall n\in\Bbb N):\frac{a_{n+1}}{a_n}=\frac3{2+a_n}.$$Therefore, if $a_n\in(0,1)$, then $\frac{a_{n+1}}{a_n}>1$, which means that $a_{n+1}>a_n$. So, $(a_n)_{n\in\Bbb N}$ is indeed increasing. But$$a_n<1\implies1-a_{n+1}=2\frac{1-a_n}{2+a_n}>0,$$and therefore $a_{n+1}<1$ too. So, $\lim_{n\to\infty}a_n=1$.
Can you deal with the case in which $a_1>1$ now?
